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frx

  • one year ago

1). \[\lim_{(x,y) \rightarrow (0,0)} \frac{ x ^{2}y ^{2} }{ x ^{4}+y ^{4} }\] 2). \[\lim_{(x,y) \rightarrow (0,0)} \frac{ x ^{2}y ^{2} }{ 2x ^{4}+y ^{4} }\] In the first limit I used that \[f(x,y)=f(x,0) \rightarrow 0\] and \[f(x,y)=f(0,y)\rightarrow0\] But in the second one that becomes totally wrong, since the x-axis goes to zero and y-axis to 1/3. Why should one change from f(o,y)-> 0 to x=y in the second one, how do I recognize when I should set x=y ?

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  1. experimentX
    • one year ago
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    *

  2. frx
    • one year ago
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    ?

  3. experimentX
    • one year ago
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    just a bookmark!!

  4. shubhamsrg
    • one year ago
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    Isn't answer to 1st one 1/2 ?

  5. frx
    • one year ago
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    Nope it's 0.

  6. experimentX
    • one year ago
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    it's a double limit, you have to be a bit careful with it ... since there are multiple ways to come at (0,0), I think there is a theorem .... but currently I am busty posting my own question.

  7. shubhamsrg
    • one year ago
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    http://www.wolframalpha.com/input/?i=limit+%28x%5E2+y%5E2+%2F%28x%5E4+%2By%5E4%29%29+as+x-%3E0+and+y-%3E0 What does this mean ?

  8. experimentX
    • one year ago
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    says that limit does not exist, try y=x or y=x^2, if you don't get the same limit then limit does not exist.

  9. frx
    • one year ago
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    According to the answerkey to Adams Calculus the answer for the first one is 0 and the second one DNE

  10. Zarkon
    • one year ago
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    \[\lim_{(x,y) \rightarrow (0,0)} \frac{ x ^{2}y ^{2} }{ x ^{4}+y ^{4} }\] let \(y=mx\) then \[\lim_{x \rightarrow 0} \frac{ x ^{2}(mx) ^{2} }{ x ^{4}+(mx) ^{4} }\] \[\lim_{x \rightarrow 0} \frac{ m^2x ^{4} }{ x ^{4}(1+m ^{4}) }\] \[\lim_{x \rightarrow 0} \frac{ m^2 }{ 1+m ^{4} }=\frac{ m^2 }{ 1+m ^{4} }\] pick any real number \(m\)

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